| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find constant from definite integral |
| Difficulty | Standard +0.3 Part (a) is straightforward differentiation and solving a linear equation. Part (b) requires standard integration of power functions and substituting limits, then solving a polynomial equation in k. Both parts use routine AS-level techniques with no novel insight required, making this slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \left[\frac{x^{-1/2}}{2k}\right] - \left[\frac{x^{-3/2}}{2}\right] + ([0])\) | B2, 1, 0 | \(([0])\) implies that more than 2 terms counts as an error |
| Sub \(\frac{dy}{dx} = 3\) when \(x = \frac{1}{4}\). Expect \(3 = \frac{1}{k} - 4\) | M1 | |
| \(k = \frac{1}{7}\) (or \(0.143\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int \frac{1}{k}x^{1/2} + x^{-1/2} + \frac{1}{k^2} = \left[\frac{2x^{3/2}}{3k}\right] + \left[2x^{1/2}\right] + \left[\frac{x}{k^2}\right]\) | B2, 1, 0 | OE |
| \(\left(\frac{2k^2}{3} + 2k + 1\right) - \left(\frac{k^2}{12} + k + \frac{1}{4}\right)\) | M1 | Apply limits \(\frac{k^2}{4} \to k^2\) to an integrated expression. Expect \(\frac{7}{12}k^2 + k + \frac{3}{4}\) |
| \(\frac{7}{12}k^2 + k + \frac{3}{4} = \frac{13}{12}\) | M1 | Equate to \(\frac{13}{12}\) and simplify to quadratic. OE, expect \(7k^2 + 12k - 4 = 0\) |
| \(k = \frac{2}{7}\) only (or 0.286) | A1 | Dependent on \((7k-2)(k+2) = 0\) or formula or completing square |
| 5 |
## Question 10(a):
| $\frac{dy}{dx} = \left[\frac{x^{-1/2}}{2k}\right] - \left[\frac{x^{-3/2}}{2}\right] + ([0])$ | B2, 1, 0 | $([0])$ implies that more than 2 terms counts as an error |
|---|---|---|
| Sub $\frac{dy}{dx} = 3$ when $x = \frac{1}{4}$. Expect $3 = \frac{1}{k} - 4$ | M1 | |
| $k = \frac{1}{7}$ (or $0.143$) | A1 | |
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int \frac{1}{k}x^{1/2} + x^{-1/2} + \frac{1}{k^2} = \left[\frac{2x^{3/2}}{3k}\right] + \left[2x^{1/2}\right] + \left[\frac{x}{k^2}\right]$ | B2, 1, 0 | OE |
| $\left(\frac{2k^2}{3} + 2k + 1\right) - \left(\frac{k^2}{12} + k + \frac{1}{4}\right)$ | M1 | Apply limits $\frac{k^2}{4} \to k^2$ to an integrated expression. Expect $\frac{7}{12}k^2 + k + \frac{3}{4}$ |
| $\frac{7}{12}k^2 + k + \frac{3}{4} = \frac{13}{12}$ | M1 | Equate to $\frac{13}{12}$ and simplify to quadratic. OE, expect $7k^2 + 12k - 4 = 0$ |
| $k = \frac{2}{7}$ only (or 0.286) | A1 | Dependent on $(7k-2)(k+2) = 0$ or formula or completing square |
| | **5** | |
---10 A curve has equation $y = \frac { 1 } { k } x ^ { \frac { 1 } { 2 } } + x ^ { - \frac { 1 } { 2 } } + \frac { 1 } { k ^ { 2 } }$ where $x > 0$ and $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item It is given that when $x = \frac { 1 } { 4 }$, the gradient of the curve is 3 .
Find the value of $k$.
\item It is given instead that $\int _ { \frac { 1 } { 4 } k ^ { 2 } } ^ { k ^ { 2 } } \left( \frac { 1 } { k } x ^ { \frac { 1 } { 2 } } + x ^ { - \frac { 1 } { 2 } } + \frac { 1 } { k ^ { 2 } } \right) \mathrm { d } x = \frac { 13 } { 12 }$.
Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q10 [9]}}